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2 years ago

Competitive inhibition occurs when a

Chemistry

2 answers:

Aleksandr-060686 [28]2 years ago

6 0

The answer to your question is C.

ioda2 years ago

3 0

Answer:

Competitive inhibition occurs when a molecule binds to an enzyme in the active site and prevents the substrate from binding. (C)

Explanation:

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A sample of an ideal gas at 1.00 atm and a volume of 1.50 L was placed in a weighted balloon and dropped into the ocean. As the

Lady bird [3.3K]

Answer:

Explanation:

Using Boyle's law

P₁ V₁ = P₂V₂ and temperature is constant

where P₁(pressure) = 1.00atm, P₂ = 25 atm, V₁( volume) = 1.50L V₂ =

V₂ = ( P₁ V₁ ) / P₂ = ( 1 atm × 1.50L ) / 25 atm = 0.06 L

6 0

3 years ago

Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel ha

ddd [48]

Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

K = 0.060

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Concentration of $PCl_5(g)$ = $\frac{numbers of moles}{volume}$

Concentration of $PCl_5(g)$ = $\frac{3.98}{10.0}$

Concentration of $PCl_5(g)$ = 0.398 moles

If we construct an ICE table for the above equation; we have:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

Initial 0.398 0 0

Change - x + x + x

Equilibrium (0.398 - x) x x

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

K = $\frac{[x][x]}{[0.398-x]}$

K = $\frac{x^2}{0.398-x}$

0.060 = $\frac{x^2}{0.398-x}$

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0 (quadratic equation)

a = 1; b= 0.06; c= -0.02388

Using quadratic formula;

= $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

= $\frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}$

= $\frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}$

= $\frac{-0.06+/-\sqrt{0.09912} }{2}$

= $\frac{-0.06+/-0.3148}{2}$

= $\frac{-0.060+0.3148}{2}$ or $\frac{-0.060-0.3148}{2}$

= $\frac{0.2548}{2}$ or $\frac{-0.3748}{2}$

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles of Cl₂ produced at equilibrium = 1.274 moles

7 0

3 years ago

What is the majority of the atom composed of *TWO WORD ANSWER* ANSWER ASAP WILL AWARD BRAINLIEST! SCAM ANSWERS WILL BE REPORTED

Olin [163]

Answer:

Most of the atom is empty space. The rest consists of a positively charged nucleus of protons and neutrons surrounded by a cloud of negatively charged electrons. The nucleus is small and dense compared with the electrons, which are the lightest charged particles in nature..

Explanation:

Hope it helps you..

Y-your welcome in advance..

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6 0

3 years ago

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel

n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

Answer: The mole fraction of gas B (hydrogen gas) is 0.557

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)